Projective modules over $kG$ equivalent to injective.

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Let $k$ be a field and $G$ is finite group. I want to prove that a $kG$ module $P$ is projective iff it's injective. I proved that if module is projective then it's injective.

1) $kG$ is injective because $Hom_{kG}(M,kG)=Hom_{k}(M,k)$. So every free $kG$ module is injective.

2) Every projective module is a free summand of free, so it's injective.

But I don't know how can I prove that injective modules are projective

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If $M$ is any $kG$-module, then $kG \bigotimes_{k} M$ is a free $kG$-module, and the kG-module homomorphism

$$ m \mapsto \sum_{g \in G} g \otimes g^{-1} m$$

is injective. If $M$ is injective, then since free $kG$-modules are injective, $M$ is now a direct summand of a free $kG$-module and is therefore projective.